A289799 p-INVERT of (n^3), where p(S) = 1 - S - S^2.

1, 10, 62, 377, 2232, 13015, 75898, 444014, 2601503, 15244128, 89303905, 523084546, 3063814838, 17945741321, 105115487400, 615706236199, 3606449444722, 21124456768934, 123734572586495, 724763983514112, 4245239506761217, 24866107799273146, 145650985218990062

Offset

0,2

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A289780 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (9, -27, 55, -36, 55, -27, 9, -1)

Formula

G.f.: (1 + x - x^2 + 34 x^3 - x^4 + x^5 + x^6)/(1 - 9 x + 27 x^2 - 55 x^3 + 36 x^4 - 55 x^5 + 27 x^6 - 9 x^7 + x^8).

a(n) = 9*a(n-1) - 27*a(n-2) + 55*a(n-3) - 36*a(n-4) + 55*a(n-5) - 27*a(n-6) + 9*a(n-7) - a(n-8).

Mathematica

z = 60; s = x*(1 + 4*x + x^2)/(1 - x)^4; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000578 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289799 *)
LinearRecurrence[{9, -27, 55, -36, 55, -27, 9, -1}, {1, 10, 62, 377, 2232, 13015, 75898, 444014}, 30] (* Harvey P. Dale, Jan 07 2024 *)

Crossrefs

Cf. A000578, A289780.

Sequence in context: A240157 A350644 A196632 * A027254 A370764 A159240

Adjacent sequences: A289796 A289797 A289798 * A289800 A289801 A289802

Keyword

nonn,easy

Author

Clark Kimberling, Aug 12 2017

Status

approved